Welcome to our community

Be a part of something great, join today!

Concave Downward Function and Average

renyikouniao

Member
Jun 1, 2013
41
Please show that if f is a continuous concave downward function on [a,b],then the average of the function f is greater than f[(a+b)/2],thank you in advance.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,191
This conclusion is false. Consider $-x^{2}+1$ on the interval $[-1,1]$. It's continuous and concave downward everywhere. Then the average value of the function is
$$ \frac{1}{1-(-1)} \int_{-1}^{1}(-x^{2}+1) \, dx= \int_{0}^{1}(-x^{2}+1) \, dx
= \frac{2}{3},$$
but
$$f\left( \frac{1+(-1)}{2}\right)= f(0) = 1 > \frac{2}{3} .$$
 

renyikouniao

Member
Jun 1, 2013
41
awesome explaination!thank you.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,701
Please show that if f is a continuous concave downward function on [a,b],then the average of the function f is greater than f[(a+b)/2],thank you in advance.
Perhaps the question should have read "Show that if f is a continuous concave downward function on [a,b],then the average of the function f is greater than [f(a)+f(b)]/2]." That is true, and you see why geometrically, if you notice that the graph of f lies above the line joining the points (a,f(a)) and (b,f(b)). Thus the area under the graph is greater than the area of the trapezium with vertices (a,0), (a,f(a)), (b,f(b)), (b,0). In other words, $$\int_a^bf(x)\,dx > \tfrac12(f(a)+f(b)(b-a),$$ from which $$\frac1{b-a}\int_a^bf(x)\,dx > \tfrac12(f(a)+f(b)).$$